Video: Discussing the Continuity of a Piecewise-Defined Function Involving Trigonometric Ratios at a Point

Discuss the continuity of the function 𝑓 at π‘₯ = 0, given 𝑓(π‘₯)= βˆ’(5 sin 8π‘₯)/(π‘₯) + 8 cos π‘₯ if π‘₯ β‰  0 and 𝑓(π‘₯) = βˆ’32 if π‘₯ = 0 .

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Video Transcript

Discuss the continuity of the function 𝑓 at π‘₯ is equal to zero, given that 𝑓 of π‘₯ is equal to negative five sin of eight π‘₯ over π‘₯ plus eight multiplied by the cos of π‘₯ if π‘₯ is not equal to zero and 𝑓 of π‘₯ is equal to negative 32 if π‘₯ is equal to zero.

The question wants us to determine the continuity or discontinuity of our piecewise function 𝑓 at the point π‘₯ is equal to zero. We recall we call a function 𝑓 continuous at the point π‘₯ is equal to π‘Ž if the following three conditions are true. First, we need that 𝑓 evaluated at π‘Ž is defined. This is the same as saying that π‘Ž is in the domain of our function 𝑓. Second, the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ must also exist. And we recall this is equivalent to saying the limit as π‘₯ approaches π‘Ž from the right of 𝑓 of π‘₯ and the limit as π‘₯ approaches π‘Ž from the left of 𝑓 of π‘₯ both exist and are equal. Finally, we must have the limit as π‘₯ approaches π‘Ž of our function 𝑓 of π‘₯ is equal to 𝑓 evaluated at π‘Ž.

So to check our piecewise function 𝑓 is continuous at π‘₯ is equal to zero, we need to check all three of these conditions are true. So we’ll set π‘Ž equal to zero and check each of these conditions in turn. We’ll start by checking that 𝑓 evaluated at zero is defined. To evaluate 𝑓 at zero, we see that our function is defined piecewise. And we see when π‘₯ is equal to zero, our function is defined to output negative 32. So 𝑓 evaluated at zero is equal to negative 32. And that means that 𝑓 evaluated at zero is defined. So our first condition is true.

So we’ll now move on to the second condition, checking that the limit as π‘₯ approaches zero of our function 𝑓 of π‘₯ exists. To do this, we need to check the limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯ and the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ both exist and are equal. We’ll start with the limit as π‘₯ approaches zero from the right. Since π‘₯ is approaching zero from the right, we have that π‘₯ is greater than zero. And we see from our piecewise definition of the function 𝑓 of π‘₯, when π‘₯ is not equal to zero, our function 𝑓 of π‘₯ is equal to negative five sin of eight π‘₯ over π‘₯ plus eight cos of π‘₯.

So since π‘₯ is approaching zero from the right, π‘₯ is never equal to zero. And from our piecewise definition of 𝑓 of π‘₯, we have that 𝑓 of π‘₯ is equal to this expression whenever π‘₯ is not equal to zero. This means their limits will be equal. We might then try direct substitution. However, since π‘₯ is approaching zero, we see that five sin of eight π‘₯ is approaching zero and π‘₯ is approaching zero. This gives us zero over zero, which is an indeterminate form. This means we’re either going to have to manipulate our limit into one we can evaluate or we’re going to need to use our standard trigonometric limit laws.

Since our expression contains the sine function divided by π‘₯, we recall that the limit as π‘₯ approaches zero of the sin of π‘₯ divided by π‘₯ is equal to one. This is a standard trigonometric limit result which we should commit to memory. However, in our limit, we have the sin of eight π‘₯ in our numerator. So we’ll replace all instances of π‘₯ with eight π‘₯ in our standard limit result. We see that if eight π‘₯ is approaching zero, then π‘₯ must be approaching zero. And we see that one-eighth is a constant, so we can take this constant outside of our limit. Finally, we can multiply both sides of this equation by eight, giving us the limit as π‘₯ approaches zero of the sin of eight π‘₯ divided by π‘₯ is equal to eight.

Now, by using the fact that a limit of a sum is equal to the sum of a limit, we can split our limit into the sum of these two limits. And since negative five and eight are constants with respect to π‘₯, we can take these coefficients outside of our limit. We’re now ready to evaluate these limits. We’ve already shown that the limit as π‘₯ approaches zero of the sin of eight π‘₯ divided by π‘₯ is equal to eight. And since the cos of π‘₯ is a standard trigonometric function, we can evaluate this by using direct substitution. Substituting π‘₯ is equal to zero gives us the cos of zero, which we can evaluate to give us one. Therefore, our limits are equal to negative five times eight, which is negative 40, plus eight times one, which is equal to eight. This simplifies to give us negative 32.

So we’ve shown that our limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯ is equal to negative 32. We now need to show that the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ is also equal to negative 32. Let’s see what would’ve happened if instead of taking the limit as π‘₯ approaches zero from the right using our working, we’d instead use the limit as π‘₯ approaches zero from the left. Well, since π‘₯ is approaching zero from the left, we now have that π‘₯ is less than zero. However, we only use the fact that π‘₯ was not equal to zero in our working. So in our left-hand limit, our function 𝑓 of π‘₯ is still equal to this expression.

In fact, none of the steps in our working specifically used the fact that we were taking a right-hand limit. Every step in our working will be exactly the same when we’re taking the left-hand limit. This means we can conclude that the limit as π‘₯ approaches zero from the left of 𝑓 of π‘₯ is also equal to negative 32. So since both the left-hand and right-hand limit exist and are equal, we’ve shown that the limit as π‘₯ approaches zero of 𝑓 of π‘₯ exists.

The final thing we need for our function 𝑓 of π‘₯ to be continuous at zero is the limit as π‘₯ approaches zero of 𝑓 of π‘₯ is equal to 𝑓 evaluated at zero. We already found both of these values when checking the first and second conditions. When we checked the first condition, we showed that 𝑓 evaluated at zero is equal to negative 32. And when we checked our second condition, we showed that the limit as π‘₯ approaches zero from the left and the limit as π‘₯ approaches zero from the right of 𝑓 of π‘₯ is equal to negative 32. This is the same as saying the limit as π‘₯ approaches zero of 𝑓 of π‘₯ is equal to negative 32. So we have the limit as π‘₯ approaches zero of 𝑓 of π‘₯ is equal to negative 32, which is also equal to 𝑓 evaluated at zero. So our third condition for continuity is also true.

Therefore, since our function 𝑓 of π‘₯ satisfied all three conditions for continuity at π‘₯ is equal to zero, we can conclude that our function 𝑓 is continuous at π‘₯ is equal to zero.

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